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A ball of mass 0.20kg hits a wall at an angle of 45^{\circ} with a velocity of 25m/s. If the ball rebound at 90^{\circ} to the direction of incidence with same speed. Calculate magnitude of change in momentum of the ball.

Option: 1

5\sqrt2Ns


Option: 2

5 Ns


Option: 3

3\sqrt2Ns


Option: 4

10Ns


Answers (1)

best_answer

Solution

Given-

mass, m=0.2kg

initial and final speed, v=25 m/s

Magnitude of Initial linear momentum (\vec{p_{i}}) and final linear momentum (\vec{p_{f}})-

p=|\vec{p_{f}}|=|\vec{p_{i}}|=mv \Rightarrow p=5 kg.m/s

Angle between initial and final momemtum vectors is 90^{0}, as shown in the figure below-

Change in linear momentum-

\Delta \vec{p}=\vec{p_{f}}-\vec{p_{i}}\\ \ \ |\Delta \vec{p}|=|\vec{p_{f}}-\vec{p_{i}}|\\ |\Delta \vec{p}| =\sqrt{p^2+p^2-2p^2cos90^0}\\ |\Delta \vec{p}|=\sqrt{2}p\\ \Rightarrow |\Delta \vec{p}|=5\sqrt{2} kg.m/s

Posted by

Ritika Jonwal

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